PSEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion
PSEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion
PSEB 9th Class Science Solutions Chapter 9 Force and Laws of Motion
→ Gallileo and Isac Newton put forth the scientific basis with regard to the motion of the objects.
→ The concept of force is based on the activity of pull, push or hit/kick.
→ By the application of force, the size and shape of an object can be changed.
→ Force is of two types:
- Balanced force
- Unbalanced force
→ An unbalanced force acting on an object produces motion in the object.
→ The frictional force acts against the direction of motion of an object.
→ The first law of motion is also called the Laws of Inertia.
→ The tendency of the body to remain in the state of rest or of uniform motion is called Inertia.
→ Everybody opposes the change in its state of motion.
→ The moment of inertia of the train is more than that of the cart so that it does not get displaced on being pushed. In other words, the heavier bodies have more inertia.
→ The inertia of an object is the measure of its mass.
→ The momentum ‘p’ of an object is measured by the product of its mass ‘m’ and velocity ‘v’. i.e. p = m × v
→ The momentum has both magnitude and velocity. Its direction is the same as that of velocity.
→ The S.I. unit of momentum is kg ms.
→ Force changes the momentum of an object.
→ To reduce the effect of friction the surface is either made plane or the surface is painted with a lubricant.
→ S.I. unit of force is the newton.
→ For the motion of the object due to the force acting on it, Newton gave three fundamental laws of motion which hold good for all types of motion.
→ According to the first law, everybody tends to remain in its state of rest or of uniform motion unless some external force acts on it.
→ According to the second law, the rate of change of momentum of a body is directly proportional to the unbalanced force acting on it and in its direction.
→ According to the third law, when one body applies force on another body, the second body also applies instantaneous force on the first body. These two forces are equal in magnitude but opposite in direction.
→ By applying force the motion can be given to a body at rest and the body in motion can be brought to rest or a change can be brought about in the same direction.
→ Force: It is an external cause that produces or tends to produce a change in its state of rest or of uniform motion.
→ One Newton Force: It is the force that produces an acceleration of 1 ms-2 in a body of mass 1 kg. It is denoted by ‘N’.
→ Balanced Forces: If a number of forces acting on an object do not produce any change in its state then these forces are called Balanced Forces. Their net result is zero force.
→ Unbalanced Forces: If the result of various forces acting on an object is not zero, then these forces are called unbalanced forces.
→ Frictional Force: It is the opposing force that comes into play when a body moves over the surface of another body.
→ Inertia: It is the property of the bodies due to which they cannot change their position of rest or of uniform motion unless some external force is applied.
→ The inertia of Rest: It is the property by virtue of which a body at rest will continue to remain at rest unless some external force is applied to bring about that change.
→ The inertia of Motion: It is the property of a body by virtue of which the body in motion, will continue to move with the same uniform speed unless some external force is applied to bring about that change.
→ Momentum: The product of mass and velocity possessed by a body is called its momentum.
→ Law of Conservation of Momentum: If no external force acts on a system of particles then the total momentum of the system remains conserved.
→ Newton’s Second Law of Motion: Tine rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.
→ Newton’s Third Law of Motion: Every action has an equal and opposite reaction.
→ Newton’s First Law of Motion: If a body is at rest, it will continue to remain at rest. And if a body is moving with a uniform velocity in a straight line it will continue to do so unless some external force is applied to bring about that change in its state.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion
Long Answer Type Questions:
Question 1.
What are the different types of Forces? Explain each with the help of example
Answer:
Types of forces. There are two types of forces:
1. Balanced forces
2. Unbalanced Forces.
1. Balanced Forces: When several forces are acting simultaneously on a body and their resultant is zero, the forces are said to be balanced forces.
In the case of balanced forces if some body is at rest then it will remain at rest and if it is moving with uniform speed then it will continue to move with the same speed, as if no force is doing any work. In this way with the effect of balanced forces. There does not take place any change in position of the body.
Balanced forces change the shape of the objects, e.g. if a rubber ball is pressed between palms by applying equal and opposite forces then the shape of the ball changes. This ball no longer remains round and instead becomes flat.
Example: In a tug of war, when both teams pull the rope with equal force, then resultant forces becomes zero. Therefore, both teams remain in their places. In this situation the forces applied by the two teams are equal and opposite so get balanced.
Condition for forces to be balanced.
Two equal forces acting in the opposite direction become balanced if they act along the same line and their magnitudes are equal.
Effect of Balanced Forces: Forces acting on any object if do not change its state of rest or its motion then these do change the shape of the object.
2. Unbalanced Forces: If the resultant of the several forces acting on a body is not zero, the forces are said to be unbalanced forces. Unbalanced forces produce change in the direction of uniform motion of the body or its state of rest.
Example:
In a tug-of-war, when one of the two teams pulls the rope with a larger force, it is able to pull the weaker team towards it. Here two forces are not balanced. Therefore, it results in the motion of the weaker team towards the larger force along the rope.
Question 2.
State and explain the Newton’s First Law of Motion.
Answer:
Newton’s First Law of Motion. This law states that “In this universe all bodies will remain in their state of rest or of uniform motion in a straight line until some external force is applied to bring about change in their state.
According to this law, motion can be divided into two parts:
1. First part says that a body at rest continues in its state of rest unless some external force is applied to bring change in its state of rest. We find a book lying on the table keeps lying in the same state unless someone applies force on it to pick it up.
2. Second part says that a body jn uniform motion continues moving in straight line path with a uniform speed unless someone applies external force to stop it. But in our daily life it appears but different.
As for example when we stop pedalling a bicycle the moving bicycle stops. After studying minutely it is found that between tyres of bicycle and ground there acts a force of friction which is an external force which opposes the motion. The resistance of air also opposes the motion of bicycle. So, moving bicycle stops moving due to these two forces.
Question 3.
What is inertia? What are its different types? Give examples for each one of them.
Answer:
Inertia: It is the property of matter by virtue ofivhich an object is unable to change by itself its state of rest or of uniform motion in a straight line.
Because of this property Newton’s First Law of Motion is also called the law of Inertia.
Types of Inertia Inertia is of three types:
1. Inertia of Rest: It means a body at rest tends to remain in its position of rest. i.e. a body at rest opposes the forces which try to move it. It can be understood clearly by the following example.
A man standing in a stationary bus or train falls backward when the bus or train suddenly starts moving forward. When the bus moves, the lower part (limbs) of his body begins to move along with the bus while his upper-part tends remain at rest due to inertia.
2. Inertia of Motion: This means a body in motion continues to move with uniform
motion in a straight line. i.e. it is the tendency of a body to remain in its state of uniform motion in a straight line.
Example: 1. A person sitting in moving bus falls forward when the moving bus suddenly stops. It is because as the bus stops the lower part of his body comes to rest along with the bus while upper part of his body continues to remain in motion due to inertia of motion.
2. An athelete runs for some distance before taking a jump so that his inertia of motion may help his muscular force to a longer jump.
3. Inertia of Direction. It is the property of a body which helps to maintain its direction i.e. it is inability of a body to change by itself its direction of motion.
Example: The mudguard is fitted in the wheel of a bicycle to protect from mud particles and water drops sticking to its wheel leaving it tangentially.
Imagine a stone tied to the end of a thread moving in a horizontal circle, while doing so, if the thread breaks then due to inertia of direction the stone flies off tangentially in a straight line.
Question 4.
State and explain Newton’s Second Law of motion with the help of this law how can we measure force.
Answer:
Newton’s Second Law of Motion. Newton’s second Law of motion helps to calculate the force required to bring a body in motion. It consists of two parts
1. The rate of change of momentum of a body is directly proportional to the applied unbalanced force and
2. the change in momentum due to the external applied force takes place in the direction of force.
i.e. According to this law “the external force applied in the body is directly proportional to the product of mass of the body and the acceleration produced in the direction of force.
Explanation: When a force acts on a body, it produces change in its momentum. If the force is doubled then the change in momentum also gets doubled. The more the force applied; the more is the change in momentum produced. Momentum is the product of mass of the body and its velocity. Generally no change in mass occurs. Therefore, the rate of change of momentum is actually the rate of change of velocity. Thus the applied force is proportional to the acceleration.
When an external force acts on a body at rest, it begins to move in the direction of force. When force acts in the direction of motion of the body then its momentum gets increased. But when force acts in a direction opposite to the direction of motion, its momentum gets reduced.
Force (F) ∝ mass (m) × acceleration (a)
or F = k × m × a ….(i)
(where k is a constant of proportionality)
If we choose the unit of force in such a way that as unit of force produces unit acceleration in a body of unit mass, then
Substituting F = 1, m = 1 and a = 1 in (1)
1 = k × 1 × 1
or k = 1
∴ from equation (i), F = 1 × m × a
or F = m × a
Force = Mass × Acceleration
Question 5.
Derive the mathematical relation for magnitude of force from Newton’s Second Law of Motion.
Answer:
Measurement of Magnitude of Force from Newton’s Law of Motion. Suppose force F acts on a body of mass’m’ for Y seconds which changes its velocity from u to υ, then
Initial momentum of body, p1 = mu
Final momentum of body, p1 = mυ
Now because final velocity (υ) is more than the initial velocity (u), therefore final momentum (p2) will be more than the initial momentum (p1) change in momentum,
p = p2 – p1
= mυ – mu
= m (υ – u)
According to second law of motion,
Question 6.
State and explain Newton’s third Law of motion.
Answer:
Newton’s third Law of Motion. This law states that “Every action has equal and opposite reaction” According to this law, there does not exist one force in isolation. Force always exists in pair i.e. forces of action and reaction always act on different bodies.
Explanation: (1) Consider two similar spring balances attached to each other through their hooks. One end of spring balance A is fastened to the fixed support. Pull the free end of spring balance B to right side. Note the readings of both the spring balances. Both will read the same as shown in fig. It is because read.
A and B pull each other in opposite direction with same force.
(2) Keep two balls A and B on a table at some distance from each other. When you push the ball. A towards B then the ball A acts on ball B. This force is represented by
FA→B According to Newton’s third law of motion, the ball B also reacts and applies
force on ball A. This force is represented as FB→A. if both the balls are similar then the magnitudes of action and reaction will be equal.
∴ FA→B = – FB→A
Action and reaction forces always act in the direction opposite to each other.
Question 7.
What is meant by the Law of Conservation of Momentum? Deduce this law mathematically with the help of Newton’s second and third law of motion.
Answer:
Law of Conservation of Momentum. For a system of bodies, the total vector sum of momenta of all the bodies due to mutual action and reaction remain unchanged so long as no external force acts on the system.”
Mathematical Derivation: Consider two rubber balls of masses m1, m2 and initial velocities u1, u2 respectively. Let these collide and their velocities after collisions be υ1 and υ2 respectively. If A applies a force F on B also for time t; B also applies a force F on A for same time t.
From Newton’s Second Law of Motion:
i.e. total momentum of balls A and B before collision
= Total momentum of balls A and B after collision
This proves the law of conservation of momentum that is the total momentum remains conserved.
Short Answer Type Questions:
Question 1.
What is force? Give its units.
Answer:
Force. Force is an agent which
- produces or tends to produce motion in the body
- stops the moving body or tends to stop
- increases the speed of the body or tries to increase the speed therefore,
Force may be defined is as physical cause (a push or a pull) which changes or tends to change the state of rest or uniform motion or direction of motion of a body. The force exerted by the engine makes the train to move from its actual position of rest while the force exerted by the brakes slows down or stops the moving train. The force exerted on the steering wheel of a car changes its direction of motion.
Force is a vector quantity.
Units of force: The unit of force depends upon the unit of mass and acceleration. S.I. unit of force is Newton and C.G.S. unit is Dyne
1N = 105 Dynes
Question 2.
Why does the horse rider fall forward when a running horse suddenly stops?
Answer:
When horse is running, both the horse and the rider are in motion. When the horse suddenly stops, the lower part of the rider and horse come in the state of rest while the upper part of the rider remains in motion so that he falls forward.
Question 3.
When a horse suddenly gallops, the rider falls backward. Why?
Answer:
The horse and the rider form one system. Initially both are at rest. When the horse suddenly gallops then the horse and the lower part of the rider come into motion in the forward direction while the upper part of the rider’s body tends to remain at rest. Therefore, the rider falls backward.
Question 4.
Why does a passenger fall forwaid when he alights from the moving bus?
Answer:
While alighting from the moving bus the passenger falls forward because when the feet of the passenger touch the ground, his lower part suddenly comes to rest while the upper part still remains in motion. In this way the passenger falls forward.
Question 5.
Define momentum of a body. Also give its units.
Answer:
Momentum. It is defined as the quantity of motion possessed by the body. It is measured by the product of the mass and velocity of the body.
Question 6.
A fast-moving bullet when hits the windowpane makes a round hole while a stone strikes and shatters it, why?
Answer:
If stone piece strikes window pane glass, it gets shattered while a fast-moving bullet when strikes the windowpane a round hole is formed. The reason is that small particle of glass around and near the hole do not move due to inertia along with the bullet and threfore, do not scatter.
Question 7.
Explain how a dirty blanket becomes dust-free if it is jerked once or twice?
Answer:
If a dirty blanket is beaten with a stick or is given a jerk then dust particles get separated from it because when we beat or give a jerk to the blanket it comes in motion but due to inertia of rest, dust particles remained at rest and get separated and the blanket becomes dust-free.
Question 8.
Why a fan continues to rotate for sometime even after it is switched off?
Answer:
When a fan is rotating, then because of inertia of motion it continues it rotation for some time even if we switch if off. Due to friction on resistance of air it comes to rest after few seconds.
Question 9.
Why does a gun recoil when a bullet is fired from the gun? Explain.
Answer:
When bullet is not fired, then gun and bullet both are at rest, thus total momentum of both together is zero. When a bullet is fired from the gun the bullet moves with very high speed in the forward direction i.e. its momentum is very high. Now according to law of conservation of momentum, total momentum should still be zero as was before the firing of the bullet. Thus to balance the momentum of bullet in the forward direction gun recoil.
Question 10.
Why a cricket player lowers his hand while taking of catch of cricket ball?
Answer:
A lot of force is needed to stop a fast moving ball. If we lower our hands while catching the ball, acceleration of the ball is decreased and we have to apply less force of stop the ball.
Question 11.
Write differences between balanced and unbalanced forces.
Answer:
Differences between balanced forces and unbalanced forces:
| Balanced forces | Unbalanced forces |
| 1. When balanced forces act simultaneously on a body then their net resultant is zero. | When unbalanced forces act simultaneously on a body then their net resultant is not zero. |
| 2. Balanced forces are acting on a body which is at rest, these can not bring it in motion. | Unbalanced forces act on a body at rest, these can bring it in motion. |
| 3. These forces cannot bring a change in speed or direction of the motion. | This force can bring change in the speed of direction of motion. |
| 4. This force can make a change in the shape of the body. | This force can not make a change in the shape of the body. |
Question 12.
Why a boatman exists a force on water with his oars in the opposite direction to move forward?
Answer:
That force, which causes motion in any direction, is the reaction of the applied force. To move the boat in forward direction boatman exerts a force on water with his oars in the opposite direction. As a reaction to this force boat moves in forward direction because action and reaction are equal and opposite to each other.
Important Formulae:
Numerical Problems (Solved):
Question 1.
What acceleration will be produced in a body of mass 3 kg, when a force of 12 N is applied?
Solution:
Here, Force (F) = 12 N
Mass (m) = 3 kg
Acceleration (a) =?
We know F = m × a
12 = 3 × a
or a = 12/3
∴ a = 4 m s-2
Question 2.
How much force will be required to produce an acceleration of 4 m s-2 in a ball of mass 6 kg?
Solution:
Mass of the ball (m) = 6 kg
Acceleration produced in the ball (a) = 4ms-2
Force (F) = ?
We know, F = m × a
= 6 × 4
Force (F) = 24 N
Question 3.
A man throws a ball of mass 0.5 kg vertically upwards with a velocity of 10 m s-1. What will be its initial momentum? What would be its momentum at the highest point of its reach?
Solution:
Mass of the ball (m) = 0.5 kg
Initial velocity of ball (u) = 10 m s-1
Final velocity of the ball (v) = 0 (At highest point the ball comes to rest)
Initial momentum of the ball = m × u
= 0.5 × 10
= 5 kg – m/s
Final momentunvof the ball = m × υ
= 0.5 × 0 = 0
Question 4.
A steam engine of mass 3 × 104 kg pulls two wagons each of mass 2 × 104 kg with an acceleration of 0.2 m s-2. Neglecting frictional force, calculate the:
1. force exerted by the engine.
2. force experienced by each wagon.
Solution:
Mass of steam engine (m1) = 3 × 104 kg
Mass of two wagons (m2) = 2 × (2 × 104 kg)
Total mass of the engine and wagons (m) = m1 + m2
= 3 × 104 + 4 × 104
= (3 + 4) × 104 kg
= 7 × 104 kg
Acceleration (a) = 0.2 ms-2
1. We know, F = m × a
= 7 × 104 × 0.2
= 1.4 × 104 N
2. Force experienced by each wagon = 1.4 × 104 N
Question 5.
A bullet of mass 20 g moving with a speed of 500 ms-1 strikes a wooden block of mass 1 kg and gets embedded in it. Find the speed with which block moves along with the bullet.
Solution:
Suppose the final velocity of the bag with bullet embedded in it is υ.
For Bullet, m1 = 20 g = 0.02 kg, u1 = 500 m s-1, υ1 = υ
For Block, m2 = 1 kg, u2 = 0, υ2 = υ
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
Question 6.
A car travelling at the speed of 108 km/h takes 4 s to stop on applying brakes. Calculate the force acting on the car after applying brakes. Total mass of the car (including passengers) is 1000 kg.
Solution:
Inital velocity of the car (u) = 108 km/h
= 108 × 5/18
= 30 ms-1
Final velocity of the car (υ) = 0 ms-1
Total mass of the car (m) = 1000kg
Time taken to stop the car (t) = 4s
Force ‘F’ due to application of brakes = mυ–u/t
= 1000kg × 0–30/4s ms-1
= – 750 kg – ms-2
= – 750 N
Negative sign shows that the force applied by the brakes is acting in a direction opposite to the direction of motion of the car.
Question 7.
Which one requires more force, a body of mass 2 kg accelerated at the rate 5 m s-2 or a body of mass 4 kg accelerated at 2 m s-2
Solution:
Given:
m1 = 2kg, a1 = 5 ms-2
m2 = 4kg, a2 = 2 ms-2
Force required for first body, F1 = m1 × a1
= 2kg × 5ms-2
Force required for second body, F2 = m2 × a2
= 4kg × 2ms-2 = 8N
F1 > F2
From this, it is clear that the first body would require more force.
Question 8.
A bullet of mass 0.03 kg is fired from a gun of mass 3 kg which leaves the muzzle of the gun write a velocity of 100 ms-1. If bullet takes 0.003 s to come out of the gun then calculate the force acting on the gun.
Solution:
Mass of the gun (m1) = 3 kg
Mass of the bullet (m2) = 0.03 kg
Before firing, both the gun and the bullet are at rest
∴ Initial velocity of the gun (u1) = 0
Initial velocity of the bullet (u2) = 0
After firing. Final velocity of the gun (υ1) =?
Final velocity of the bullet (υ2) = 100 m s-1
According to the law of conservation of momentum
Negative sign shows that the gun moves backward and experiences a force of 1000 N
Question 9.
From a rifle of mass 5000 g a bullet of 20 g is fired with a velocity of 500 ms-1 with respect to the ground. Find the velocity of recoil of the rifle.
Solution:
According to law of conservation of momentum,
MV + mυ = 0
V = – mυ/M
Now m = 0.02 kg, υ = 500 ms-1 and M = 5 kg
∴ 20×500/5000
or V = – 2ms-1
Negative sign shows recoiling of the rifle.
Question 10.
A girl of 40 kg mass jumps with a horizontal velocity of 5 ms-1 on a stationary trolley. The wheels of the skate are frictionless. What will be the velocity of the girl in the position of start of the trolley. Suppose no unbalanced force is acting in the horizontal direction.
Solution:
Suppose in the initial motion of the trolley, the velocity of the trolley and the girl is υ
Total initial momentum of the girl and trolley = 40 kg × 5 m s-1 + 3 kg × 0 ms-1
= 200 kg ms-1 + 0
= 200 kg – ms-1
Total momentum of the girl and trolley after she jumps on the trolley.
= 40 kg × υ ms-1 + 3 kg × υ m s-1
= (40 + 3) × υ kg – ms-1
= 43 υ kg – ms-1
According to the law of conservation of momentum.
43υ = 200
υ = 200/43
= 4.65 ms-1
The girl boarding on the trolley will move with a velocity of 4.65 m s-1 in the direction of jump.
Question 11.
The following is the distance-time table of an object in motion.
| Time in seconds | Distance in metres |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a) What conclusion can you draw about acceleration? Is it constant, increasing, decreasing or zero?
(b) What do vou infer about forces acting on obiect?
Solution:
(a) Table shows that the motion is accelerated and acceleration is increasing uniformly with time i.e. acceleration is increasing by 6 ms-2 every second.
(b) Since acceleration is increasing uniformly, the force is also increasing uniformly with time.
Question 12.
Two persons manage to push a motor car of mass 1,200 kg at uniform velocity on the road. The same motor can be pushed by three persons to produce an acceleration of 0.2 ms-2. With what force does each person push the motor car? Assume that all persons push motor car with same muscular effort.
Solution:
Here, mass of the car (m) = 1200 kg
Acceleration produced in the car (a) = 0.2 ms-2
Acceleration produced by first two persons in the car = 0
It is clear that when third person pushes the car, an unbalanced force acts on the car which produces an acceleration of 0.2 ms-2
∴ Force applied by three persons together (F) = m × a
= 1200 × 0.2 = 240 N
Now because three persons together push the car using their muscular force to produce an acceleration of 0.2 ms-2
∴ Push (Force) applied by each person = F/3
= 240N/3
= 80 N
Question 13.
A hammer of mass 500 g moving at 50 ms-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
Question 14.
A motorcar of mass 1,200 kg is moving along a straight line with uniform velocity of 90 km h-1. Its velocity is slowed down to 18 km h-1 in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
Here, mass of the car (m) = 1200 kg
Time (t) = 4 s
Initial velocity of the car (u) = 90 km/h
Very Short Answer Type Questions:
Question 1.
To bring a body into motion, what is required to be done?
Answer:
It is required to be pulled, pushed or kicked.
Question 2.
Why does an object fall down?
Answer:
Due to unbalanced gravitational force.
Question 3.
Which type of force is required to change the direction of motion of the body – a balanced or unbalanced force?
Answer:
An unbalanced force is required.
Question 4.
Why does a body stop after rolling down for some time?
Answer:
Due to frictional force.
Question 5.
How can force of friction be reduced?
Answer:
By polishing/smearing the surface with a lubricant.
Question 6.
Which scientist postulated the three laws of motion?
Answer:
Newton.
Question 7.
By which other name the first law of motion is known?
Answer:
Law of Inertia.
Question 8.
Of heavy and light objects, which have more inertia?
Answer:
Heavier objects have more inertia.
Question 9.
What is the S.I unit of momentum?
Answer:
The S.I. unit of momentum is kg – ms-1.
Question 10.
Why is talcom powder sprinkled on carrom board while playing?
Answer:
In order to reduce friction.
Question 11.
Why does an athelete run before taking a high jump?
Answer:
To increase inertia in order to take high leap.
Question 12.
What is law of conservation of momentum?
Answer:
Law of conservation of momentum. The sum total of momentum of two objects before and after collision remains same unless some external force is applied.
Question 13.
A bus and a ball are moving with the same speed. To stop which one would require more force?
Answer:
Due to more inertia of bus, more force is required to be applied for stopping it.
Question 14.
A vehicle stops on applying brakes. During this activity, what happens to its momentum?
Answer:
In this activity the major part of momentum of the vehicle is transferred to the ground while the remaining part is transferred to the air molecules.
Question 15.
1 kg wt is equal to how many newtons?
Answer:
1 kg wt = 9.8 Newtons.
Question 16.
1 newton is equal to how many kg wt?
Answer:
1 newton = 0.102 kg wt.
Question 17.
On which physical quantity inertia of an object depends?
Answer:
On mass of the object.
Question 18.
On which Newton’s law of motion, the working principle of rocket is based?
Answer:
On Newton’s third law of motion.
Science Guide for Class 9 PSEB Force and Laws of Motion InText Questions and Answers
Question 1.
Which of the following has more inertia:
(a) rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?
Answer:
We know that mass of an object is the measure of its inertia. The more is the mass of an object, the more is its inertia hence.
(a) a stone of the same size has more inertia than a rubber ball.
(b) a train has more inertia than a bicycle.
(c) a ₹ 5 coin has more inertia than ₹ 1 coin because a ₹ 5 coin has more mass than a ₹ 1 coin.
Question 2.
In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal-keeper. The goal-keeper of opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.
Answer:
To push, to strike or to pull, all these activities act as a force for changing the velocity or for changing the direction of motion of the object. Therefore, in the above given example the velocity of ball changes four times.
- First time the football player of first-team kicks the football to another player of his team and thus changes the velocity of the football.
- In second time velocity changes when the second player kicks the football towards the goal-keeper of the opposite team and applies force on the ball.
- Third time the goal-keeper pushes the ball and reduces its velocity to zero by applying force.
- The goal-keeper now applies a force by kicking the football towards player of his team. In this case the force increases the velocity of the football.
Question 3.
Explain why some of the leaves may get detached from the tree if we vigorously shake its branch.
Answer:
Before shaking, the branch of the tree, both the branch and leaves were at rest. When we shake the branch of the tree, branch moves but the leaves remain at rest due to inertia of rest and get detached from the branch.
Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and falls backward when it accelerates from rest?
Answer:
When the bus is moving, whole of our body is moving forward. When brakes are applied, the lower part of our body touching the bus (e.g., feet etc.) comes to rest and upper part of the body not touching the bus continue move forward due to inertia of motion and fall in forward direction.
When the bus suddenly starts and accelerates from rest, the lower part of our body starts moving forward (accelerating) along with the bus while upper part of our body tends to remain at rest due to inertia of rest and we fall backward.
Question 5.
If action is always equal to reaction, explain how a horse can pull a cart?
Answer:
According to Newton’s third law of motion “Action and Reaction are equal and opposite.”
The horse pulls (Action) the cart with some force in the forward direction and the cart applies equal force on the cart in the backward direction (Reaction). These two forces balance each other. When the horse pushes the ground with its feet in the backward direction with force P along OP it gets reaction R due to ground along OR in the upward direction.
This force of reaction can be resolved into two rectangular components.
1. Vertical component ‘V’ which balances the weight mg of the horse and cart in the downward direction.
The horizontal component ‘H’ which helps to move the cart in the forward direction. The force of friction between wheels and ground acts in the backward direction but the horizontal component ‘H’ acts in the forward direction is more than the backward force of friction, it succeeds to move the cart forward.
Question 6.
Explain why is it difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity?
Answer:
Water is ejected from rubber hose in forward direction with a force (action), it exerts an equal reaction on the hose in backward direction. Due to backward reaction, fire man finds it difficult to hold the hose.
Question 7.
From a rifle of mass 4 kg, a bullet df> mass 50 g is fired with an initial velocity of 35 m s-1. Calculate the recoil velocity of the rifle.
Solution:
Mass of the bullet (m1) = 50 g = 0.05 kg
Mass of the rifle (m2) = 4 kg
Initial velocity of the bullet (u1) = 0
Initial velocity of the rifle (u2) = 0
Final velocity of the bullet (υ1) = 35 m s-1
Final velocity of the rifle (υ2) = ?
According to the law of conservation of momentum,
Total initial momentum of bullet and rifle = Total final momentum of the bullet and rifle.
∴ Negative sign indicates that the rifle moves in a direction opposite to the direction of motion of the bullet.
Question 8.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms-1 and 1ms-1 respectively. They collide and after the collision, the first object moves with a velocity of 1.67 m s-1. Determine the velocity of the second object.
Solution:
PSEB 9th Class Science Guide Force and Laws of Motion Textbook Questions and Answers
Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, it is possible for an object to move with non-zero velocity even when net zero unbalanced force is experienced by it. In this situation the magnitude of velocity and direction will be same. As for example, in case of a rain drop falling freely with constant velocity, the weight of the drop is balanced by upthrust so to say the net unbalanced force on drop is zero.
Question 2.
When a carpet is beaten with a stick, the dust comes out of it? Explain.
Answer:
When a carpet is beaten with a stick, the carpet is set into motion while the dust particles due to inertia tend to remain at rest. In this way dust particles get detached from the carpet and come out of it.
Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with rope?
Answer:
When a fast-moving bus suddenly takes a turn round a sharp bend then the luggage placed on the roof of a bus gets displaced. The reason for this is that the luggage tends to remain with linear motion while an unbalanced force is applied by the engine to change the direction of the bus so that the luggage kept at the roof of the bus gets displaced. So it is advised to tie the luggage with a rope on the roof of bus.
Question 4.
A batsman hits a cricket ball when then rolls on a level ground. After covering short distance, the ball comes to rest. The ball comes to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball so that ball would want to come to rest.
Answer:
(c) is correct. There is a force of friction on the ball in direction opposite to that of motion.
Question 5.
A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force on it if its mass is 7 metric tonnes. (1 tonne = 100 kg)
Solution:
Question 6.
A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
Here, mass of stone (m) = 1 kg
Initial velocity of stone (u) = 20 ms-1
Distance travelled by the stone (S) = 50 m
Final velocity of stone (υ) = 0 (at rest)
Force of friction between the stone and ice (F) = ?
Using υ2 – u2 = 2aS, we have
(0)2 – (20)2 = 2 × (a) × 50
– 20 × 20 = 100 × a
a = −20×20/100
a = – 4 ms-2
F = ma = 1 × (- 4)
F = – 4 N
Minus sign shows the force of friction is in direction opposite to direction of motion of stone.
Question 7.
A 8,000 kg engine pulls a train of 5 wagons, each of 2,000 kg along a horizontal track. If the engine exerts a force of 40,000 N and track offers a force of friction of 35,000 N, then calculate the
(a) net accelerating force;
(b) acceleration of the train; and
(c) force of wagon 1 on wagon 2.
Solution:
Mass of the engine = 8000 kg
Mass of 5 wagons = 5 × 2000 kg = 10,000 kg
∴ Total mass of engine and 5 wagons = 8000 kg + 10,000 kg = 18,000 kg
Total force of engine = 40,000 N
Frictional force offered by the track= 5000 N
(a) ∴ Net Accelerating Force (F) = Total force of engine – Frictional force of track
= 40,000 N – 5000 N = 35,000 N
(c) Force exerted by wagon 1 on wagon 2 = Net Accelerating Force – Mass of wagon × Acceleration
Question 8.
An automobile vehicle has mass of 1,500 kg. What must be the force between vehicle and the road if vehicle is to be stopped with negative acceleration of 1.7 ms-2?
Solution:
Here the mass of automobile (m) = 1500 kg
Acceleration of vehicle (a) = -1.7ms-2
Frictional Force between road and vehicle (F) = ?
We know, F = m x a
= 15000 x (- 1.7)
= – 2550 N
∴ Backward frictional force (F) = 2550 N
Question 9.
What is the momentum of an object of mass m moving with velocity υ?
(a) (mυ)2; (b) mυ2; (c) 1/2 mυ2, (d) mυ.
Answer:
(d) is correct. Momentum = mυ.
Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor with constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
When no acceleration is to be produced (i.e. body is to be moved with constant velocity), the net force has to be zero. Force of friction should be equal and opposite to the force applied i.e., force of friction has to be 200 N.
Question 11.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in the opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of combined object after collision?
Solution:
Given, mass of the lirst object (m1) – 1.5 kg
, and mass of the second object (m2) = m1 = 1.5 kg
Initial velocity of the first object (u1) = 2.5 ms-1
Initial velocity of the second object (u2) = – 2.5 ms-1
(Since both the objects move in the direction opposite to each other therefore velocity of first object will be taken as positive and that of the other as negative.)
Suppose after collision the velocity of the combination of two objects is ‘υ’
∴ According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
m1u1 + m2u2 = m1υ + m2υ
1.5 × 2.5 + 1.5 × (-2.5) = (1.5 × υ + 1.5 × υ)
1.5 [2.5 + (-2.5)] = (1.5 + 1.5) × υ
1.5 [2.5 – 2.5] = 3 × υ
1.5 × 0 = 3 × υ
0 = 3 × υ
∴ υ = 0 ms-1
i. e. Both the objects will come to rest after collision.
Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the road side, it will probably not move. A student justifies by answering that the two forces cancel each other. Comment on the logic and explain why the truck does not move.
Answer:
Student is justified. Friction is equal and opposite to force applied till the force applied crosses the force of limiting friction. When he applies a force slightly more than force of limiting friction, the truck will move. Till the truck moves uniformly, the force applied is exactly equal to force of friction at that instant.
Question 13.
A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the change in momentum occurred in the motion of hockey ball by the force applied by hockey stick.
Sol. Mass of the ball (m) = 200 g
= 200/1000 kg = 0.2 kg
Initial velocity of the ball (u) = 10 ms-1
Final velocity of the ball (v) = – 5ms-1
[∵ the direction of the ball is opposite to the first direction]
Change in momentum of the ball = Final momentum – Initial momentum.
= mυ – mu
= m (υ – u)
= 0.2 (- 5 – 10)
= 0.2 × (- 15)
= – 3.0 kg – ms-1
Question 14.
A bullet of mass 10 kg travelling horizontally with a velocity of 150 ms-1 strikes a stationary wooden block and come to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Calculate the magnitude of force exerted by the wood in block in the bullet.
Solution:
Question 15.
An object of mass 1 kg travelling in straight line with a velocity of 10 ms-1 collides with it and sticks to a stationary wooden block of mass 5 kg. Then both move off together in the same straight line. Calculate the total momentum before the impact and just after the impact. Also calculate the velocity of combined object.
Solution:
Mass of the object (m1) = 1 kg
Initial velocity of the object (u1) = 10 ms-1
Mass of the wooden block (m2) = 5 kg
Initial velocity of the wooden block (u2) = 0 [Wooden block at rest]
Suppose ‘υ’ is the velocity of the combination of the object and wooden block after collision
∴ Before collision total momentum of the object and block
= m1u1 + m2u2
= 1 × 10 + 5 × 0
= 10 + 0
= 10kg ms-1 ……………… (i)
After collision. Total momentum of the object and block = m1υ + m2υ
= (m1 + m2) × υ
= (1 + 5) × υ
= 6υ kg m – ms ……… (ii)
According to the law of conservation of momentum,
Total momentum of the combinations before collision = Total momentum of the combination after collision 10 = 6υ
υ = 10/6
∴ υ = 5/3 ms-1
= 1.67 ms-1
Substituting the value of υ in (ii) above
∴ Total momentum of the combination (after collision) = 6υ
= 6 × 5/3
= 10kg – ms-1
Question 16.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6s. Calculate the initial and final momentum of the object. Also find the force exerted on the object.
Solution:
Question 17.
Akhtar, Kiran and Rahul were riding in a motor car that was moving with a high velocity on an express-way when an insect hit the windshield and got struck on wind-screen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of motor car (because change in the velocity of insect was much more than that of motor car). Akhtar said that since the motor car was moving with a larger velocity, it exerted a larger force on the insect. As a result, the insect died. Rahul while putting in entirely new explanation said that both the motor car and the insect experienced the same force and same change in their momentum. Comment on these suggestions.
Answer:
I agree with Rahul’s explanation. According to law of conservation of momentum, during collision, the momentum of the system (insect and motor car) remains conserved. Therefore, both insect and motor car experience the same force and hence same change in momentum. The insect having smaller mass would suffer greater change in velocity as a result of this, it will crush the insect while the motor car does not suffer any noticeable change in velocity.
Question 18.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms-2.
Solution:
Here, momentum of the dumb-bell (m) = 10 kg
Initial velocity of the bell (u) = 0 (at rest)
Distance covered by the bell (S) = (h) = 80 cm
= 0.80 m
Acceleration of the ball (a) = 10 ms-2 (downward direction)
Let υ be the final velocity of the bell on reaching the ground.
Using υ2 – u2 = 2aS
υ2 – (0)2 = 2 × 10 × 0.80
υ2 = 2 × 10 × 0.80
or υ2 = 16
∴ Final velocity of the bell (υ) = √16 = 4ms
Momentum transferred by the bell to the floor (p) = m × υ
= 10 × 4 = 40 kg – ms-1
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